Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2017 / S3 / 1

The Bohr model assumes Coulomb attraction and classical circular motion between transitions; selected stationary orbits do not radiate; their angular momentum is $mvr=n\hslash$; and a transition exchanges a photon of energy $h\nu =E_i-E_f$. The first two dynamical equations giveHere $n=1,2,\dots$, $m$ is the electron mass under the fixed-proton approximation, $e$ its charge magnitude, and ${ϵ}_0$ the vacuum permittivity. The stationary-orbit postulate deliberately suspends the classical radiation prediction; this is a historical model, not a consistent deduction from classical electrodynamics.

Substitute the quantized radius in $mvr=n\hslash$ to get $v_n=e^2/(4\pi {ϵ}_0\hslash n)$. Coulomb potential energy is $-e^2/(4\pi {ϵ}_0{r}_n)$ and the stipulated virial relation makes the total one half of this value. HenceThe negative sign means a bound state relative to a separated stationary proton and electron.

The emitted photon has energy $E_3-E_2=\mathrm{Ry}(1/4-1/9)=5\mathrm{Ry}/36$. ThereforeThis is the Balmer-alpha transition in this approximation. Absorption in the reverse direction has the same photon energy; fine structure and radiative corrections are beyond the model.

Separate center-of-mass motion and replace $m$ by the reduced mass $\mu =mM/(m+M)$ in the relative-coordinate formulas. Then $r_n$ is the proton-electron separation, energy levels and the Rydberg constant scale by $\mu /m$, and transition wavelengths increase by $m/\mu$. The electron's own orbital radius about the center of mass is $M{r}_n/(M+m)$; the proton radius is $m{r}_n/(M+m)$. The relative speed retains $\alpha c/n$, while the electron's individual speed acquires $M/(M+m)$. Replacing the mass in the radius alone but leaving energies unchanged would be inconsistent.